Find $\int \dfrac{1}{2x^2-4x+20}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac16 \text{arcsin}\left(\dfrac{x-1}{3}\right)+C$ (Choice B) B $\dfrac13 \text{arctan}\left(\dfrac{x-1}{3}\right)+C$ (Choice C) C $\dfrac13 \text{arcsin}\left(\dfrac{x-1}{3}\right)+C$ (Choice D) D $\dfrac16 \text{arctan}\left(\dfrac{x-1}{3}\right)+C$
Explanation: The integrand is in the form $\dfrac{1}{p(x)}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as a constant multiple of $(x+h)^2+k^2$. Then, we will be able to integrate using the following formula, which is based on the derivative of the inverse tangent function: $\int \dfrac{1}{(x+ h)^2+ k^2}\,dx=\dfrac{1}{ k} \text{arctan}\left(\dfrac{x+ h}{ k}\right)+C$ [Why is this formula true?] We start by rewriting $p(x)$ as a constant multiple of $(x+ h)^2+ k^2$ : $\begin{aligned} 2x^2-4x+20&=2[x^2-2x+10] \\\\ &=2[x^2-2x+1+9] \\\\ &=2[(x-1)^2+9] \\\\ &=2[(x{-1})^2+{3}^2] \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{2x^2-4x+20}\,dx \\\\ &=\int\dfrac{1}{2[(x-1)^2+3^2]}\,dx \\\\ &=\dfrac12\int\dfrac{1}{(x{-1})^2+{3}^2}\,dx \\\\ &=\dfrac12\cdot\dfrac{1}{{3}} \text{arctan}\left(\dfrac{x{-1}}{{3}}\right)+C \\\\ &=\dfrac16 \text{arctan}\left(\dfrac{x-1}{3}\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{2x^2-4x+20}\,dx=\dfrac16 \text{arctan}\left(\dfrac{x-1}{3}\right)+C$